When designing a resonant circuit, for example as a local oscillator, the product of $L$ and $C$ determine the resonant frequency according to $$f_{res} = \frac{1}{2\pi \sqrt{L C}}.$$

This means that we can reach a frequency of $7030 \text{ kHz}$ for example with

  • $L = 2.33 ~\mu H$ and $C = 220~pF$, or
  • $L = 23.3 ~\mu H$ and $C = 22~pF$, or
  • $L = 513 ~nH$ and $C = 1~nF$, etc.

If we increase the inductance, the quality factor of the circuit should increase, since $$Q = \frac{X_L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}.$$

Does the quality of the $LC$ circuit continuously increase with increasing $L$, or is there a maximum that can be reached at some $L$-to-$C$ ratio? How do I find this ratio and wit this pick the best combination of $L$ and $C$ for my circuit?

  • 1
    Do these answers over on Electronics answer your question? electronics.stackexchange.com/questions/201028/… – Kevin Reid AG6YO Nov 25 at 21:49
  • @KevinReidAG6YO My question is basically building onto that answer. It says there that a further increase in the inductance improves the Q factor, until the self-resonant frequency of the inductor is reached. It doesn't go into (practical) details as to how the "practically best" combination of $L$ and $C$ can be found. Is there a formula or some rule to estimate how large the capacitor compared to the inductor should be? – ahemmetter Nov 25 at 21:57
  • Worthy of note is that the formula given for Q is for series and not parallel tuned circuits – Scott Earle Nov 26 at 2:20
up vote 7 down vote accepted

There is no hard and fast rule. Consider that in some LC circuit applications, a lower Q may be desirable in order to achieve a wider bandwidth. In other cases, a very high Q may be desirable for narrow selectivity, for example.

Both the inductor and the capacitor in a resonant circuit may affect the Q of the circuit.

The Q of the inductor is determined by its inductive reactance divided by its series resistance.

$$Q_L=\frac{X_L}{R_L} \tag1$$

Since inductance is generally a factor of turns squared and since the resistance of the inductor is a factor of turns, this first order analysis indicates that the lower the inductor value for a given wire type/diameter and for a given construction method, the higher the Q of the inductor. The higher the Q of the inductor, the higher the Q of the resonant circuit.

The Q of a capacitor is determined by its capacitive reactance divided by its effective series resistance.

$$Q_C=\frac{X_C}{ESR_C} \tag2$$

In most practical cases, the ESR of the capacitor is a factor only in series resonant circuits. In a parallel resonant circuit, generally the series resistance of the inductor will dominate the Q.

In a series resonant circuit, the resistive losses of the inductor and capacitor are simply added. Since you quoted the formula for a series resonant circuit, this should be your approach.

$$Q=\frac{1}{R_L+ESR_C} \sqrt{\frac{L}{C}} \tag3$$

Before selecting a final inductor value, make certain that its self resonance will not negatively affect your circuit.

The insertion loss under this scenario is given as:

$$ \text{Insertion Loss} = 20\log\left({1-\frac{Q}{Q_L}}\right) \tag 4$$

where QL is as noted above and Q is the series circuit Q as noted in the equation in your question.

Other factors that may come into play, depending upon the application, are the inductance and the Q of the inductor over a wide frequency range; the stability of the capacitor over a wide temperature range; and the size, weight, tolerance, cost and availability of the components.

  • Also, availability of components play a big role, as do tolerances, and available space. Note that the frequency-dependent behaviour of capacitors is different from the frequency-dependent behaviour of inductors, so especially for wideband applications, you start simulating things to figure out what your circuit really does - admittedly, a resonant circuit is the opposite of a "wideband application", but I'd guess the same as for frequency applies for temperature... Analog is hard. That's why I like DSP so much: if the math works, it works. – Marcus Müller Nov 26 at 10:48

There is definitely a best L to C ratio. You write: For 7030 kHz for example L=2.33 μH and C=220 pF, or L=23.3 μH and C=22 pF, or L=513 nH and C=1 nF, etc. If we increase the inductance, we can use a thinner wire for the same unloaded quality factor in all those cases.

Assume the unloaded Q=300 in all the cases. If you make a parallel resonator in a 50 ohm feed line the loaded Q and therefore the bandwidth will be very different in the different cases. The impedance at resonance for Q=300 is (maybe approximately, I used a nomogram) L=2.33 μH and C=220 pF, R=3000 ohms loaded Q=5, loss=1.6%, bw=1.4 MHz or L=23.3 μH and C=22 pF, R=30000 ohms loaded Q=1(?) loss=0, bw=80 MHz or L=513 nH and C=1 nF, R=660 ohms loaded Q=23 loss=8% bw=300 kHz etc.

An unloaded LC resonator can be regarded as a series or parallel resonator. Just two components. It is the way you connect it to something else that determines which kind it is. Unloaded Q is the same.

If you connect something in series with L and C (now I am doing qualified guessing) with Q=300 you should find this in a 50 ohm system on resonance: L=2.33 μH and C=220 pF, R=0.3 ohms or L=23.3 μH and C=22 pF, R=3 ohms or L=513 nH and C=1 nF, R=0.066 ohms.

A resonator that has an unloaded to loaded Q ratio above 10 will attenuate by less than 10%. (0.4 dB) If the unloaded to loaded Q=2, the loss would be 3 dB and the bandwidth given by the loaded Q. Parallel or series the same.

One can also use an LC resonator as series and parallel at the same time. Inject a current into the resonator and take out the voltage across it. Yoy can find one example here: https://www.youtube.com/watch?v=BgZUuX0tzn8

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